AND xor OR


Problem Statement :


Given an array  of  distinct elements. Let  and  be the smallest and the next smallest element in the interval  where .

.

where , are the bitwise operators ,  and  respectively.
Your task is to find the maximum possible value of .

Input Format

First line contains integer N.
Second line contains N integers, representing elements of the array A[] .


Output Format

Print the value of maximum possible value of Si.



Solution :



title-img


                            Solution in C :

In C ++ :





#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;


int main() {
   
    
    
    int n;
    cin>>n;
    
    int * arr;
    
    
    arr =  new int[n];
    
    
    for(int i=0;i<n;i++)cin>>arr[i];
    
    stack<int> S;
    
    int m = 0;
    
    for(int i=0;i<n;i++)
    {
        while(true)
        {
            if(S.empty())
            {
                S.push(arr[i]);
                break;
            }
            
            else if(S.top() < arr[i] )
            {
                m = max(m, S.top()^arr[i]);
                S.push(arr[i]);
                break;
            }
            else
            {
                S.pop();
            }
        }        
    }
    
    
    stack<int> S2;
    
    
    for(int i=n-1;i>=0;i--)
    {
        while(true)
        {
            if(S.empty())
            {
                S.push(arr[i]);
                break;
            }
            
            else if(S.top() < arr[i] )
            {
                m = max(m, S.top()^arr[i]);
                S.push(arr[i]);
                break;
            }
            else
            {
                S.pop();
            }
        }        
    }
    
    
    
    cout<<m<<endl;
    return 0;
}








In Java :




import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] arr = new int[n];
        for(int i = 0; i < n; i++){
            arr[i] = in.nextInt();
        }
        
        Stack<Integer> stack = new Stack<Integer>();
        int result = 0;
        for(int i = 0; i < n; i++){
            while(true){
                if(stack.isEmpty()){
                    stack.add(arr[i]);
                    break;
                } else if(stack.peek() < arr[i]){
                    result = Math.max(result, stack.peek() ^ arr[i]);
                    stack.add(arr[i]);
                    break;
                } else{
                    stack.pop();
                }
            }
        }
        stack = new Stack<Integer>();
        
        for(int i = n - 1; i >= 0; i--){
            while(true){
                if(stack.isEmpty()){
                    stack.add(arr[i]);
                    break;
                } else if(stack.peek() < arr[i]){
                    result = Math.max(result, stack.peek() ^ arr[i]);
                    stack.add(arr[i]);
                    break;
                } else{
                    stack.pop();
                }
            }
        }
        
        System.out.println(result);        
    }
}








In C :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

long long int sol(int A, int B)
{
	return (((A & B) ^ (A | B)) & (A ^ B));
}

long long int inter(long long int arr[], int n)
{
	int i,j;
	long long int min1 = 9999999999, min2 = 9999999999, max = -1;
	for(i=0;i<n-1;i++)
	{
			if(sol(arr[i], arr[i+1]) > max)
				max = sol(arr[i], arr[i+1]);
        if(i+2 <= n-1)
            if(arr[i+1] > arr[i] && arr[i+1] > arr[i+2] && sol(arr[i], arr[i+2]) > max)
                max = sol(arr[i], arr[i+2]);
		
	}
	return max;
	
}

int main()
{
    int n,i;
	scanf("%d", &n);
	long long int arr[n];
	for(i=0;i<n;i++)
	{
		scanf("%lli", &arr[i]);
	}
		printf("%lli", inter(arr, n));
    return 0;
}








In  Python3 :





n_elements = int(input())
data = list(map(int, input().split()))
assert len(data) == n_elements

res = 0
minima = [data[0]]
for i in range(1, n_elements):
    d = data[i]
    while len(minima):
        x = minima.pop()
        res = max(d ^ x, res)
        if x < d:
            minima.append(x)
            break
    minima.append(d)
print(res)
                        








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